Results 1 to 12 of 12

Math Help - Theorem tennis match!

  1. #1
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1

    Theorem tennis match!

    Okay, so this is a non-standard problem. The game is to play tennis with theorems. Gameplay goes as follows :

    (1) A theorem comes your way.
    (2) You respond to the theorem with another theorem, which you must relate to the first.

    If you play, try to keep the game open, by stating theorems which can easily be related to by others.

    I'll begin with a well-known theorem from high-school geometry : the medians of a triangle intersect.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Quote Originally Posted by Bruno J. View Post
    Okay, so this is a non-standard problem. The game is to play tennis with theorems. Gameplay goes as follows :

    (1) A theorem comes your way.
    (2) You respond to the theorem with another theorem, which you must relate to the first.

    If you play, try to keep the game open, by stating theorems which can easily be related to by others.

    I'll begin with a well-known theorem from high-school geometry : the medians of a triangle intersect.

    I'm not sure I completely grasp the goal of such a game, but let's play a little:

    Theorem: The intersection point of the three medians in any triangle divides each median in a proportion 1:2, where the largest part is always on the vertex's side.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    If X, Y, Z are identically distributed real-valued random variables, then \frac{E(X)}{E(X+Y+Z)}:\frac{E(Y+Z)}{E(X+Y+Z)} = 1/2.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Bruno J. View Post
    If X, Y, Z are identically distributed real-valued random variables, then \frac{E(X)}{E(X+Y+Z)}:\frac{E(Y+Z)}{E(X+Y+Z)} = 1/2.
    Hmm you have to make asumptions about integrability, and about the fact that E[X+Y+Z] is not 0.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Quote Originally Posted by Moo View Post
    Hmm you have to make asumptions about integrability, and about the fact that E[X+Y+Z] is not 0.


    Not to mention that it'd be nice to know how this relates to the first two geometric theorems, though I can guess a probable rather tenuous link...

    Tonio
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
    Posts
    722
    Quote Originally Posted by tonio View Post
    Not to mention that it'd be nice to know how this relates to the first two geometric theorems, though I can guess a probable rather tenuous link...

    Tonio
    I'm guessing perhaps he meant you can use parts of the previously posted theorem in your one.

    Like if I said

    The cardinality of the integers is \aleph_0

    Your next theorem could be about cardinality, the integers or something to do with \aleph_0.

    Is that what's going on in here..?

    (Although that does seem a little simple for a maths challenge question...)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Quote Originally Posted by Moo View Post
    Hmm you have to make asumptions about integrability, and about the fact that E[X+Y+Z] is not 0.
    Absolutely right! Make all necessary assumptions.

    Tonio : here's how I thought it related to the geometric theorem. The point of intersection of the medians is the center of gravity of the three vertices, assuming they have equal mass. With this in mind, the theorem you stated is a specific case of the following more general theorem : the center of mass of m+n points of equal (nonzero ) mass, n of which are blue and m red, lies n/m of the way along the segment joining the centers of mass of the red points and of the blue points. But we can also view the points as events in a real probability space, with the expectation replacing the center of mass...

    Deadstar : that's essentially it. The point is just to have fun and to see what kind of unexpected theorems people will pull out of their hats.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Quote Originally Posted by Bruno J. View Post
    Absolutely right! Make all necessary assumptions.

    Tonio : here's how I thought it related to the geometric theorem. The point of intersection of the medians is the center of gravity of the three vertices, assuming they have equal mass. With this in mind, the theorem you stated is a specific case of the following more general theorem : the center of mass of m+n points of equal (nonzero ) mass, n of which are blue and m red, lies n/m of the way along the segment joining the centers of mass of the red points and of the blue points. But we can also view the points as events in a real probability space, with the expectation replacing the center of mass...

    Deadstar : that's essentially it. The point is just to have fun and to see what kind of unexpected theorems people will pull out of their hats.


    Hmmm...center of mass smell big time to physics mingling into our mathematics. Don't like that

    Would the game perhaps be more interesting if we stick to only maths?

    Tonio
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Maybe the expression "center of mass" can equivalently be replaced by "centroid", or "barycenter" ? I think it doesn't change the true meaning of what Bruno says, and it's a real mathematical notion !
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Quote Originally Posted by Moo View Post
    Maybe the expression "center of mass" can equivalently be replaced by "centroid", or "barycenter" ? I think it doesn't change the true meaning of what Bruno says, and it's a real mathematical notion !
    Haha! What do you guys have against physics?

    Come on guys, the ball is in your court. Feel free to reset the thread if you are not inspired by the current discussion!
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor chiph588@'s Avatar
    Joined
    Sep 2008
    From
    Champaign, Illinois
    Posts
    1,163
    Quote Originally Posted by Bruno J. View Post
    Haha! What do you guys have against physics?

    Come on guys, the ball is in your court. Feel free to reset the thread if you are not inspired by the current discussion!
    Ok, I'll start with a new theorem.

    Let  A be a complex  n\times n matrix. Let  R_i=\sum_{j\neq i}|a_{ij}| and  D(a_{ii},R_i) be the closed disc centered at  a_{ii} and radius  R_i .

    Then each eigenvalue of  A lies in at least one  D(a_{ii},R_i) .
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Alright!

    This one is from Putnam a few years back. Let B be a finite collection of open discs in the plane (possibly overlapping). Then there is a subcollection B' of them, such that they are pairwise disjoint, and such that \displaystyle\cup_{D \in B'}3D covers \cup_{D \in B}D, where 3D denotes the disc with the same center as D but with three times the radius.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. tennis / height ?
    Posted in the Calculus Forum
    Replies: 0
    Last Post: March 1st 2011, 03:18 PM
  2. [SOLVED] tennis / height ?
    Posted in the Trigonometry Forum
    Replies: 0
    Last Post: March 1st 2011, 03:08 PM
  3. Tennis Balls
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 8th 2009, 10:44 AM
  4. tennis
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: August 26th 2009, 02:17 PM
  5. Tennis probability
    Posted in the Statistics Forum
    Replies: 2
    Last Post: April 14th 2009, 09:10 PM

Search Tags


/mathhelpforum @mathhelpforum